# COVID-19 NZ: Can one person infect 2.7 others?

In this post we investigate to see if one person with COVID-19 can infect 2.7 other people on average.

Note:
R, Ro, and Re are used for the number of people than can be infected by one person (the number of new cases infected by one person).
Ro (called R-zero or R-nought) is used when there is no isolation.
Re is used when there is isolation.
R is used for general discussion.

Below is an updated table from my previous post:
https://aaamazingphoenix.wordpress.com/2020/05/22/covid-19-nz-would-the-number-of-cases-be-double-if-l4-delayed-one-week/
(model starts with a total of 10 cases- extra numbers backfilled) While the values in the first row highlighted in red above do not match, they are closer than in the previous post and can be considered zero after rounding.

We are only considering values from 16 March anyway (from when we have an estimated total of 10 cases) but can consider 11 March as well.

In the above table R = 5.3782.

R = 5.3782 would appear very  high. It would suggest that one person could on average infect about 5.4 others.

We consider other values for R. We note that if R> 1 then infection is spreading and if R < 1 then the infection is diminishing and the outbreak is likely to end.

We look at the table below:

 n R = 5.3782/n 1 5.37824 2 2.68912 3 1.792747 4 1.34456 5 1.075648 6 0.896373

The right hand column is 5.37824 divided by the left column.

We note that alongside 6, R  < 1 (and hence can be ignored).

We consider the values for R for 2 (2.68912) and 5 (1.075648).

It is possible that after an incubation period, a person may infect people for the next 5 days. On the next day, another person could also infect people for five days, and so on.

This means that on each day, people can be infected by people first showing symptoms (symptomatic) / able to infect others on that day and from people symptomatic from each of the previous four days. i.e. The value for R comes from 5 infected people.

This means that the R-value 5.37824 is five times higher than the actual value for R, 5.37824 / 5 = 1.075648

However this would mean that one person would only infect about 1.08 others.

We consider this number too low.

However it is possible that one person can infect others for two cycles.

This could give us an R-Value (for 2 cycles = 10 days) of

(1.4^10)/10
(since there would be 10 people infecting others on each day for two cycles)

=(5.37824^2) / 10

= 28.9254655/10

= 2.89254655

~ 2.9

We now consider the another scenario.

We have seen that it may be possible for a person to infect others not just for one cycle (5 days after the incubation period) but for two cycles (10 days).

This would mean that for each day, the value for R (5.37824) is a combination of people infected from the current cycle and also from the previous cycle (the second infection cycle for the previous group).

This suggests that the value for R is twice the actual value. i.e.

R = 5.37824 / 2

= 2.68912

We note that this value is close to 2.7 and conclude that this is may be a suitable value for Ro or Re.

The above values were calculated using a value of a = 1.4 (see previous post).

Using a = 1.41, the value for R is

(1.41 ^ 5)/2 = 2.786542

This means that 2.8 is also a possible value for Ro or Re.

Since our our curve calculated using a = 1.4  fits the Actual cases numbers well, we adopt 2.7 for Re up until 26 March (first full day of Lockdown Level 4).

We also conclude that Ro is at least 2.7 (and from our above discussion may be up to about 2.9) since isolation (even if only self-isolation) was in effect before Lockdown Level 4 came into effect at 11.59 pm on 25 March.

We conclude there is likely to be an incubation period of one cycle (5 days) and an infectious period of two cycles (10 days).

Below are graphs from my previous post:  My other COVID-19 posts can be found here:
https://aaamazingphoenix.wordpress.com/tag/coronavirus/

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